Advertisements
Advertisements
Question
Find n, if (n + 3)! = 110 × (n + 1)!
Advertisements
Solution
(n + 3)! = 110 × (n + 1)!
∴ (n + 3)(n + 2)(n + 1)! = 110 × (n + 1)!
∴ (n + 3)(n + 2) = 110
∴ n2 + 5n + 6 = 110
∴ n2 + 5n + 6 − 110 = 0
∴ n2 + 5n − 104 = 0
∴ n2 + 13n − 8n − 104 = 0
∴ n(n + 13) − 8(n + 13) = 0
∴ (n + 13)(n − 8) = 0
∴ n + 13 = 0 or n − 8 = 0
∴ n = − 13 or n = 8
But n ∈ N
∴ n ≠ − 13
Hence, n = 8.
APPEARS IN
RELATED QUESTIONS
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
A teacher wants to select the class monitor in a class of 30 boys and 20 girls, in how many ways can the monitor be selected if the monitor must be a boy? What is the answer if the monitor must be a girl?
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Evaluate: 6!
Compute: `(12/6)!`
Compute: (3 × 2)!
Compute: `(9!)/(3! 6!)`
Compute: `(8!)/(6! - 4!)`
Compute: `(8!)/((6 - 4)!)`
Write in terms of factorial:
3 × 6 × 9 × 12 × 15
Write in terms of factorial:
5 × 10 × 15 × 20 × 25
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 8, r = 6
Find n, if `"n"/(6!) = 4/(8!) + 3/(6!)`
Find n if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
Show that
`("n"!)/("r"!("n" - "r")!) + ("n"!)/(("r" - 1)!("n" - "r" + 1)!) = (("n" + 1)!)/("r"!("n" - "r" + 1)!`
Find the value of: `(8! + 5(4!))/(4! - 12)`
Find the value of: `(5(26!) + (27!))/(4(27!) - 8(26!)`
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
